Question

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B,3% buy B and C and 4% buy A and C. If 2% families buy all the three news papers, find the number of families which buyAA onlyBB onlyC None of A, B and C.

Answer 1

Number of families$=10000$

$n\left(A\right)=\frac{40\times 10000}{100}=4000$

$n\left(B\right)=\frac{20\times 10000}{100}=2000$

$n\left(C\right)=\frac{10\times 10000}{100}=1000$

$n(A\cap B)=\frac{5\times 10000}{100}=500$

$n(A\cap C)=\frac{4\times 10000}{100}=400$

$n(B\cap C)=\frac{3\times 10000}{100}=300$

$n(A\cap B\cap C)=\frac{2\times 10000}{100}=200$

no. of people buy only $A$, only $B$ & only $C=1400+200+3300=4900$